I tried to solve this generally so I could just plug in n later for any other n's that I may need, but somewhere I may have messed up because my coefficients ended up all being zero for the n=1 series... So, all I need is someone to see if I messed up...

I started off by noting that the equation was in general form and that p(x) and q(x) are both analytical everywhere so x = 0 is a valid point for a convergent series solution.

Next (and this part is where I had my doubts because it is an ordinary point, not a singular one), I took the limits of xp(x) and and got zero for both. This lead to the roots:

and

Now, I prepared a generalized solution:

Plugging this into the DE yields:

When the x in the second term is distributed, the second and third sum can be added together:

Next I multiplied through by to eliminate negative indices:

Then I shifted the indices, the first sum gets the replacement k=i the second one gets k=i+2:

Then I took out the first two terms of the first sum:

Then I combined the two sums:

Then I set the coefficients equal to 0 and used r = 1:

-> is arbitrary.

->

Now, when n=1, every coefficient is zero except so the solution for n=1 is:

When n=2, All of the odd numbered coefficients are zero so that the odd number powers remain, and I get:

Thanks in advance for any help.

Feb 20th 2010, 09:47 AM

HallsofIvy

That equation is NOT singular at any point and you do not need to use
Frobenius' method. Just use a solution of the form .